Short Loop Polygons and Quantitative Energy Decrease — Examples

The actual midpoint side. Represent the vertices by unit vectors $u_0,u_1,u_2$ with $u_i\cdot u_j=c:=\cos s\in (0,1)$ for $i\mathrm{\ne }j$. The midpoint of the short arc from $u_i$ to $u_j$ is $(u_i+u_j)\mathrm{/}\sqrt{2+2c}$. Two adjacent such midpoints have dot product $(u_i+u_j)\cdot (u_j+u_k)\mathrm{/}(2+2c)=(1+3c)\mathrm{/}(2+2c)$, so their spherical distance is $s_1=\arccos ((1+3c)\mathrm{/}(2+2c))$. The rotational symmetry fixes the north pole and permutes these midpoints, so the midpoint triangle is equilateral with that same centre.

Strict contraction. For $0<c<1$, the quotient $u=(1+3c)\mathrm{/}(2+2c)$ satisfies $u<1$ and $u-c=(1-c)(1+2c)\mathrm{/}(2+2c)>0$. Thus $0<s_1=\arccos u<\arccos c=s$ by [L3].

The midpoint triangle is equilateral (i). The symmetry group of the equilateral triangle acts transitively on its vertices and fixes the centre $N$; the midpoint operation is equivariant under isometries of $S^2$, so $fx$ is again an equilateral triangle with centre $N$ and side $s_1$; hence $L(fx)=3s_1$, $E(fx)=3s_1^2$ and $\mathrm{mesh}(fx)=s_1$, and step 2.1 gives the strict inequalities of (i).

The iteration contracts to zero (ii). By step 3.1 applied to each iterate, $f^{k}x$ is equilateral with side $s_k$, where $s_{k+1}=\arccos ((1+3\cos s_k)\mathrm{/}(2+2\cos s_k))$, and $(s_k)$ is strictly decreasing and bounded below by $0$; hence it converges to some $\lambda \ge 0$ by A monotone sequence converges if and only if it is bounded, and continuity of the recursion ([L1], [L3]) gives $c_{\mathrm{\infty }}=(1+3c_{\mathrm{\infty }})\mathrm{/}(2+2c_{\mathrm{\infty }})$ with $c_{\mathrm{\infty }}=\cos \lambda >0$, so $(c_{\mathrm{\infty }}-1)(2c_{\mathrm{\infty }}+1)=0$ and $c_{\mathrm{\infty }}=1$, hence $\lambda =0$. Moreover $1-\cos s_{k+1}=(1-\cos s_k)\mathrm{/}(2+2\cos s_k)\le (1-\cos s_k)\mathrm{/}2$, so $\sin (s_k\mathrm{/}2)\le 2^{-k\mathrm{/}2}\sin (s_0\mathrm{/}2)$. Since $s_k\le s_0<\pi \mathrm{/}2$, The limit of sin x divided by x at zero is one bounds $u\mathrm{/}\sin u$ near zero; away from zero its numerator is bounded and its denominator is bounded below by [L3]. Thus $s_k\le C{2}^{-k\mathrm{/}2}$ for some $C>0$, establishing the geometric rate in the title. Therefore $L(f^{k}x)=3s_k\to 0$, so $x\in C_h^0(3)$ by the description of the basin [L2]. Moreover $L(f^{k}x)=3s_k\in (0,2\pi )$ for every $k$, so the uniform decrement of [L5] applies to each iterate and gives $E(f^{k+1}x)\le E(f^{k}x)-{\lambda }_3(3s_k)$, in agreement with the strict drop computed in step 3.1.

The equality case does not occur (iii). By step 3.1, $E(fx)=3s_1^2<3s^2=E(x)$ for every $s>0$, so the tuple is neither constant nor equally spaced along a closed local geodesic; the deficit $E(x)-E(fx)$ is strictly positive and the equality characterization of L2 does not apply.

Convergence to the constant tuple. The three vertex vectors have equal dot product with the north-pole vector $N$ and sum to a positive multiple of $N$. Thus ${\cos }^2{\rho }_k=(1+2\cos s_k)\mathrm{/}3$, by squaring their sum; since $s_k\to 0$, ${\rho }_k\to 0$; hence the iterates converge uniformly to the constant tuple at the north pole.

Conclusion. Clause (i) is steps 1.1, 2.1 and 3.1, clause (ii) is steps 4.1 and 5.1, and clause (iii) is step 4.2: the midpoint iteration on a small equilateral spherical triangle stays equilateral, contracts the side strictly, converges to the centre and realises a strict energy drop at every step.

Constant tuples (i). If $x=(p,\dots ,p)$ then every consecutive distance is $0$, so $\mathrm{mesh}(x)=L(x)=E(x)=0$; every pair $(x_i,x_{i+1})$ is degenerate with midpoint $p$ by [L1], so $f(x)=x$, and then $L(f^{k}x)=0$ for all $k$, so $x\in C_h^0(n)$ by [L2]. The corresponding constant loop is short (length $0<2\pi$) and shrinkable through constant loops, and conversely $E(x)=0$ forces all edges to have length $0$, i.e. $x$ to be constant.

Degeneracy at zero (ii). The chosen positive minorant obeys $0<{\lambda }_n(2r)\le r^2\mathrm{/}(2n^4)$, so ${\lambda }_n(2r)\to 0$ as $r\to 0^{+}$. No boundedness assertion about $\delta$ near a parameter-domain endpoint is needed.

Degeneracy at $2\pi$ (ii). As $r\to {\pi }^{-}$, $\eta =(\pi -r)\mathrm{/}2\to 0$ and $\mu =\min \{r\mathrm{/}(2n),\eta \}\to 0$. The intrinsic separation construction gives $0<\delta (\eta ,\mu )\le 2(\mu \mathrm{/}4)=\mu \mathrm{/}2$, since its cut chord length is nonnegative. Therefore $0<{\lambda }_n(2r)\le \delta r\mathrm{/}(2n)\le \mu r\mathrm{/}(4n)\to 0$. This proves the claimed limit using an actual bound, rather than continuity at a point outside $\delta$'s domain.

Collapsed edges (iii). A zero edge contributes zero to mesh, length and energy, and its midpoint is its endpoint by [L1]. If $x$ is nonconstant, another edge is positive and $E(x)>0$; equality of energies cannot hold, since [L2] requires all positive-length equality edges to have the same positive length, contradicting the zero edge. When $x$ is in the short basin, put $r=L(x)\mathrm{/}2>0$ and $\mathrm{\Delta }=E(x)-E(fx)$. The variance estimate of [L3] gives $\mathrm{\mid }{\xi }_i-2r\mathrm{/}n\mathrm{\mid }\le n\sqrt{\mathrm{\Delta }}$; at the zero edge this forces $\mathrm{\Delta }\ge 4r^2\mathrm{/}{n}^4>r^2\mathrm{/}{n}^4$, so the first decrement case applies. The perturbation supplier is needed instead for degenerate face triples in the small-deficit case, where all edges are $\ge r\mathrm{/}n>0$. Finally, deleting consecutive repeated vertices preserves the normalized polygonal loop; if both lists have at least three entries, $X$ is geodesic and their common length is $<2\pi$, [L7] identifies both basin memberships with shrinkability of that same loop. Equal initial energies alone do not identify their different midpoint iterations.

The basin criterion (ii), (iv). The explicit ${\lambda }_n$ supplies no uniform positive decrement near either length endpoint, by steps 1.2 and 1.3. The basin's defining condition is ${\lim }_{k}L(f^{k}x)=0$, and L2 gives the equivalent eventual threshold $L(f^{m}x)<l\mathrm{/}2$. Openness follows from that strict threshold; closedness inside $\{L<2\pi \}$ uses [L3]'s bounded iteration on positive compact length bands. Neither conclusion requires a positive lower bound for ${\lambda }_n$ near zero.

Conclusion. Clause (i) is step 1.1, clause (ii) is steps 1.2, 1.3 and 2.1, clause (iii) is step 1.4, and clause (iv) is step 2.1: the zero-length boundary is a genuine limit of the basin, the decrement function degenerates at both endpoints of $(0,2\pi )$, and degenerate edges do not disturb the estimates; AC enters only through the suppliers [L3] and [L4] ([L6]).

The data of the tuple (i). Consecutive points $j\mathrm{\ell }\mathrm{/}n$ and $(j+1)\mathrm{\ell }\mathrm{/}n$ are at distance $\mathrm{\ell }\mathrm{/}n$ (the shorter arc has length $\mathrm{\ell }\mathrm{/}n<\mathrm{\ell }\mathrm{/}2$, so it is the unique geodesic), so $\mathrm{mesh}(x)=\mathrm{\ell }\mathrm{/}n\le h$ and $x\in P_h(n)$, $L(x)=n\cdot \mathrm{\ell }\mathrm{/}n=\mathrm{\ell }$ and $E(x)=n(\mathrm{\ell }\mathrm{/}n{)}^2={\mathrm{\ell }}^2\mathrm{/}n$.

The midpoint tuple is the shifted tuple (i). The arclength midpoint of the arc from $j\mathrm{\ell }\mathrm{/}n$ to $(j+1)\mathrm{\ell }\mathrm{/}n$ is $(j+1\mathrm{/}2)\mathrm{\ell }\mathrm{/}n$, so $f(x{)}_j=(j+1\mathrm{/}2)\mathrm{\ell }\mathrm{/}n$; this is the equally spaced tuple based at $\mathrm{\ell }\mathrm{/}(2n)$, i.e. a rotation of $x$ by $\mathrm{\ell }\mathrm{/}(2n)$. The rotation is an isometry of $S_{\mathrm{\ell }}^1$, so $L(f^{k}x)=\mathrm{\ell }$ and $E(f^{k}x)={\mathrm{\ell }}^2\mathrm{/}n$ for every $k$; in particular the iterated lengths do not tend to $0$ and $x\mathrm{\notin }{C}_h^0(n)$.

Straightness and the equality case (ii). Every consecutive triple of $x$ lies on the unique minimizing arc of length $2\mathrm{\ell }\mathrm{/}n<\mathrm{\ell }\mathrm{/}2$ because $n\ge 5$, and its middle point is the midpoint of that arc; hence each triple is straight, which is exactly the equality condition of L2: the tuple is the list of $n$ equally spaced points of the closed local geodesic $S_{\mathrm{\ell }}^1$. Since the basin contains no nonconstant straight equilateral tuple by The uniform energy decrement on the basin, bounded iteration, and the closedness of the basin inside the short polygon space (iii), this is consistent with step 2.1.

The short circle and its minimum (iii). By [L1] the circle is compact locally CAT(1), but not CAT(1). Pairs at distance $<\mathrm{\ell }\mathrm{/}2$ have a unique shortest arc. If an isometrically embedded circle has length $u$, its opposite points have two distinct minimizing arcs of length $u\mathrm{/}2$, so $u\mathrm{/}2\ge \mathrm{\ell }\mathrm{/}2$. The whole circle realizes equality, proving $m=\mathrm{\ell }$. By L3 it is nonshrinkable, consistent with the basin-only decrement and the constant positive iterated energy of step 2.1.

The boundary case (iv). For $\mathrm{\ell }=2\pi$ the same computations give $\mathrm{mesh}(x)=2\pi \mathrm{/}n$, $L(x)=2\pi$, $E(x)=4{\pi }^2\mathrm{/}n$ and $f(x)$ the shifted equispaced tuple, so its length and energy are stationary while $S_{2\pi }^1$ is CAT(1); stationary length and energy therefore do not distinguish the two cases.

Conclusion. Clause (i) is steps 1.1 and 2.1, clause (ii) is step 3.1, clause (iii) is step 3.2 and clause (iv) is step 3.3; AC enters only through the suppliers [L2] and [L3] ([L5]).

The two-sphere (i). $S^2$ is CAT(1) by [L1], so by [L3] every isometrically embedded circle in $S^2$ has length at least $2\pi$, i.e. $m\ge 2\pi$; the equator is an isometrically embedded circle of length $2\pi$, so $m=2\pi$, and L3 gives that every loop of length $<2\pi$ is shrinkable.

The latitude homotopy (ii). Parametrize the parallel by $E_{\phi }(t)=(\cos \phi \cos (2\pi t),\cos \phi \sin (2\pi t),\sin \phi )$. For an angular increment $u$ with $\mathrm{\mid }u\mathrm{\mid }\le \pi$, dot products and the addition formulas give the spherical distance $D_{\phi }(u)=2\arcsin (\cos \phi \sin (\mathrm{\mid }u\mathrm{\mid }\mathrm{/}2))$. Its right derivative at zero is $\cos \phi$, by The derivatives of sine and cosine are cosine and minus sine and For $-1<y<1$, $(\arcsin y{)}^{\mathrm{\prime }}=1\mathrm{/}\sqrt{1-y^2}$ and $(\arccos y{)}^{\mathrm{\prime }}=-1\mathrm{/}\sqrt{1-y^2}$. Consequently, for every $ϵ>0$, all sufficiently small increments satisfy $(\cos \phi -ϵ)\mathrm{\mid }u\mathrm{\mid }\le D_{\phi }(u)\le (\cos \phi +ϵ)\mathrm{\mid }u\mathrm{\mid }$. Refining any partition and summing proves that its supremum length on an angular interval of size $A$ is $A\cos \phi$, using the metric partition definition Length in a metric target: lower semicontinuity and arc-length reparametrization. Thus $L(E_{\phi })=2\pi \cos \phi$ and these loops are normalized, including the constant pole. The displayed coordinates vary uniformly continuously with $\phi$, and their lengths vary continuously. This gives the asserted null-homotopy, with short members for $\phi >0$, while $E_0$ has length $2\pi$ and is outside the domain of short-loop homotopy.

The short circle (iii). In $S_{\mathrm{\ell }}^1$, every pair at distance $<\mathrm{\ell }\mathrm{/}2$ has exactly one shortest arc, whereas opposite points have two distinct minimizing arcs of length $\mathrm{\ell }\mathrm{/}2$. Any isometrically embedded circle of length $u$ has two minimizing arcs between its opposite points, so $u\mathrm{/}2\ge \mathrm{\ell }\mathrm{/}2$. The full circle realizes equality, proving $m=\mathrm{\ell }$. By [L3] every loop of length $<\mathrm{\ell }$ is shrinkable; the full circle is nonshrinkable. For the ordinary null-homotopy assertions, lift a loop to $\mathbb{R}$ under $t\mapsto t<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>\mathrm{\ell }$: subdivision into arcs lying in intervals of length $<\mathrm{\ell }\mathrm{/}2$ gives successive unique local lifts once the initial value is fixed. The endpoint displacement is $k\mathrm{\ell }$ for an integer $k$. A loop of length $<\mathrm{\ell }$ has $\mathrm{\mid }k\mathrm{\mid }\mathrm{\ell }\le L<\mathrm{\ell }$, so $k=0$ and its lift is closed; For any loop with $k=0$, multiplying its closed lift about its initial point by $t\in [0,1]$ contracts it through loops of lengths $tL$: local lifts preserve length by the partition definition, and Euclidean scaling multiplies length by $t$. For a normalized loop this family is normalized and continuous in the uniform-plus-length topology, including $t=0$. Thus every short zero-winding loop is shrinkable. For a continuous homotopy, compact uniform continuity [L4] gives a common finite subdivision into the same local lifting charts near each parameter value; compatible local lifts therefore depend continuously on the parameter. Their endpoint displacement is a continuous integer multiple of $\mathrm{\ell }$, hence constant on the parameter interval. The full circle has $k=1$ and a constant loop has $k=0$, so the full circle is not null-homotopic. Thus every nonzero-winding short loop is nonshrinkable, has length at least $\mathrm{\ell }$, and is not null-homotopic, proving the asserted classification.

Separation (iv). In a compact geodesic locally CAT(1) space the existence of a short nonshrinkable loop is equivalent to the failure of CAT(1) by L3, and when $m<2\pi$ the minimum nonshrinkable length is $m$, realized by an isometrically embedded circle; both examples above are instances of this criterion, since $S^2$ has $m=2\pi$ and no short nonshrinkable loop, while $S_{\mathrm{\ell }}^1$ has $m=\mathrm{\ell }<2\pi$ and the full circle as shortest nonshrinkable loop.

Conclusion. Clause (i) is step 1.1, clause (ii) is step 1.2, clause (iii) is step 1.3 and clause (iv) is step 1.4; AC enters only through the supplier [L3] ([L6]).